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science chemistry

Answer 1

Answer #1

ice initial temperature = 0oC

0^oC -----------------------------------> 37^oC

(1) (2)

Q = Q1 + Q2

Q = (Cp x m )ice + (m Cp dT )water

Q = (79.7 cal / g) x ( 1.0 g ) + (1.0 g ) (1.0 cal / g. oC ) x (37 oC)

Q = 79.7 + 37

Q = 116.7 cal

heat = 117 cal

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Answer 2

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